Let $(a_n)$ a sequence such that $a_n>0 \ \forall n\in \mathbf{N}$. Suppose that $\sum_{n=0}^{\infty}a_n$ converges and the sequence $(b_n)$ is bounded. Show that $\sum_{n=0}^{\infty}a_nb_n$ converges.
I would like to have a feedback on my proof and to know if everything holds. Thanks in advance.
Proof.
First of all, as $\sum_{n=0}^{\infty}a_n$ converges, we can note that $\lim_{n\to \infty}a_n=0$. So, $\exists N \ \forall n\ge N$: $0<a_n<1$ .
Then, as $(b_n)$ is a bounded sequence, we have by definitiot that:
$\exists M>0: \ |b_n|\le M \ \forall n\in \mathbf{N}$.
With all these preliminary results we can write that:
$|b_n|\le M \iff |b_n|\cdot a_n\le M\cdot a_n \underbrace{\iff}_{a_n>0 \ \forall n} |b_na_n|\le M|a_n|$.
As $\sum_{n=0}^{\infty}a_n$ converges and $a_n>0 \ \forall n \in \mathbf{N}$ (so there is no difference between $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}|a_n|$), $\sum_{n=0}^{\infty}|a_n|$ converges and so $\sum_{n=0}^{\infty}M|a_n|$ too.
By comparaison we conclude that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely so it converges.
Your answer is correct. You can use Cauchy criterion.
Assume $ M>0$. Given $ \epsilon>0$,
For $ m $ and $ n $ large enough $$|\sum_{k=n}^ma_k|=\sum_{k=n}^ma_k< \frac{\epsilon}{M}$$ and $$|\sum_{k=n}^ma_kb_k|\le M\sum_{k=n}^ma_k<\epsilon$$
This proves the convergence of the series $\sum a_nb_n$.